A space homeomorphic to the connected sum $\mathbb{RP}^3$ # $\mathbb{RP}^3$

Question

Problem (1) Consider the space $Y$ obtained from $S^2 \times [0,1]$ by identifying $(x,0)$ with $(-x,0)$ and also identifying $(x,1)$ with $(-x,1)$, for all $x\in S^2$.Show that $Y$ is homeomorphic to the connected sum $ \mathbb{RP}^3 $#$ \mathbb{RP}^3$.

(2) Show that $S^2 \times S^1$ is a double cover of the connected sum $\mathbb{RP}^3$ # $ \mathbb{RP}^3$.

My main difficulty is that I cannot explain my proof and write down a rigor proof.

As for (1), if we decrease one dimension i.e.$ \mathbb{RP}^3 $ replaced by $ \mathbb{RP}^2 $, I can explain it with some pictures.

However, in the case of (1), I cannot use any intuition to organize my proof. I also have same problem in (2). In fact, this is a qualifying exam problem and I hope I can write down a good answer for a test purpose.

Thank you!

Answer 1

$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Proj}{\mathbf{P}}$Sketch: The space obtained from $S^{2} \times [0, 1]$ by identifying $(x, 1) \sim (-x, 1)$ and identifying $S^{2} \times \{0\}$ to a point is (pretty clearly) $\Reals\Proj^{3}$, viewed as a $3$-ball with antipodal identification on the boundary. Consequently, the space obtained from $S^{2} \times [0, 1]$ by identifying $(-x, 1) \sim (x, 1)$ is $\Reals\Proj^{3}$ with an open ball removed.

To express $S^{2} \times S^{1}$ as a double cover of $\Reals\Proj^{3} \mathop{\#} \Reals\Proj^{3}$, let $x$ denote a unit vector in $\Reals^{3}$, so $x \mapsto -x$ is the antipodal map of $S^{2}$; and let $\theta$ denote a real number in $[-\pi, \pi]$ viewed as an angular coordinate on $S^{1}$, so $\theta \mapsto -\theta$ is "reflection" of the circle.

The map $f:S^{2} \times S^{1} \to S^{2} \times S^{1}$ defined by $f(x, \theta) = (-x, -\theta)$ is a fixed-point free involutory homeomorphism. Projection on the second factor induces a map $p:(S^{2} \times S^{1})/f \to [-1, 1]$; the preimage of each endpoint is $\Reals\Proj^{2}$, and the preimage of each interior point is $S^{2}$. From this and the answer to Question 1, it's easy to see $(S^{2} \times S^{1})/f$ is homeomorphic to $\Reals\Proj^{3} \mathop{\#} \Reals\Proj^{3}$.

Answer 2

Think of $S^2$ as the unit sphere in $\mathbb{R}^3$. Then define an embedding$$\varphi:S^2\times\left[0,\frac{1}{2}\right]\hookrightarrow\mathbb{R}^4,\quad((x,y,z),t)\mapsto(x,y,z,t).$$Let $\pi:\mathbb{R}^4\to\mathbb{R}\mathbb{P}^3$ be the quotient. Then $\pi\circ\varphi$ is a quotient map from $S^2\times[0,1/2]$ to $\mathbb{R}\mathbb{P}^3$ with an open ball removed. This should give you (1).